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Mathematical Biosciences

journal homepage: www.elsevier.com/locate/mbs

Homeostasis despite instability

W. Duncana, J. Bestb, M. Golubitskyb, H.F. Nijhoutc, M. Reed⁎,a

a Department of Mathematics, Duke University, Durham, NC 27708, USAbDepartment of Mathematics, The Ohio State University, Columbus, OH 43210, USAc Department of Biology, Duke University, Durham, NC 27708, USA

A R T I C L E I N F O

Keywords:BiochemistryHomeostasisChair curveInstabilityFeedback inhibition

A B S T R A C T

We have shown previously that different homeostatic mechanisms in biochemistry create input-output curveswith a “chair” shape. At equilibrium, for intermediate values of a parameter (often an input), a variable, Z,changes very little (the homeostatic plateau), but for low and high values of the parameter, Z changes rapidly(escape from homeostasis). In all cases previously studied, the steady state was stable for each value of the inputparameter. Here we show that, for the feedback inhibition motif, stability may be lost through a Hopf bifurcationon the homeostatic plateau and then regained by another Hopf bifurcation. If the limit cycle oscillations arerelatively small in the unstable interval, then the variable Z maintains homeostasis despite the instability. Weshow that the existence of an input interval in which there are oscillations, the length of the interval, and the sizeof the oscillations depend in interesting and complicated ways on the properties of the inhibition function, f, thelength of the chain, and the size of a leakage parameter.

1. Introduction

Most human and animal physiological systems can function well inthe face of gene polymorphisms and changes in the environment be-cause of rich networks of regulatory mechanisms. This is true at alllevels of organization, from gene networks, to cellular biochemistry, totissue and organ properties, and to the behavior of the whole organism.In the simplest case, this is illustrated conceptually by the function Z(I)in Fig. 1. The variable Z depends on I, which could represent an input,an environmental variable, or some other variable of the system.However, over a large range of I (between H1 = 6 and H2 = 15 inFig. 1), the value of Z(I) changes very little. The homeostatic region isthe range of I where Z(I) changes very little. Outside the homeostaticregion the regulatory mechanisms fail and Z(I) changes rapidly as Ivaries; we call this escape from homeostasis [1]. Overall the curvegiving the dependence looks like a chair [2], and examples of such chaircurves abound: body temperature as a function of environmental tem-perature [1]; liver homocysteine concentration as a function of me-thionine input [1]; afferent arterial flow in the kidney as a function ofblood pressure [3]; cerebral blood flow as a function of cerebral bloodpressure [4]. In studying metabolism, we have found that many dif-ferent kinds of biochemical mechanisms can produce chair curves [5].Golubitsky and Stewart have used singularity theory as an analyticaltool to find nodes in networks that are homeostatic with respect to aninput [6].

We have investigated [5] many different biochemical mechanismsthat give rise to homeostasis where, typically, I is an input to the net-work. In all of these cases, the system had a unique stable equilibriumfor each I. In this paper we show an entirely new phenomenon. Thesteady state of the system can lose its stability, yet the system still showsa homeostatic region and a chair curve. Consider the simple biochem-ical chain pictured in Fig. 2. The last element in the chain, Xn, inhibitsthe reaction that takes X1 to X2. This kind of feedback inhibition is oneof the most common homeostatic mechanisms in biochemistry [5]. Thehomeostatic variable is Xn because as I increases, Xn tends to increase,which increases the inhibition by f, limiting how much Xn rises.

For =n 4 and an appropriate choice of the inhibitory function, f, thesystem shows the behavior indicated in Fig. 3. For I small, the equili-brium is stable and X4(I) increases linearly in I. At the value I1 there is aHopf bifurcation and the equilibrium becomes unstable but showshomeostasis (the red curve). Finally at I2, the equilibrium becomesstable again and shortly thereafter X4(I) shows escape from homeostasisby rising linearly with I. For I in the interval (I1, I2), the system has astable limit cycle; the green curves show the maximum and minimumvalues of X4(I, t) as X(I, t) traverses the limit cycle for fixed I.

Consider biological experiments on this system where the experi-menter chooses an input, I, and then measures X4. When I< I1, themeasurements will cluster tightly about the stable equilibria on the bluecurve. For I1< I< I2, there will be more spread of the measured valuesbecause X4 is changing in time because the dynamics has a limit cycle.

https://doi.org/10.1016/j.mbs.2018.03.025Received 8 December 2017; Received in revised form 23 March 2018; Accepted 24 March 2018

⁎ Corresponding author.E-mail address: reed@math.duke.edu (M. Reed).

Mathematical Biosciences 300 (2018) 130–137

Available online 27 March 20180025-5564/ © 2018 Elsevier Inc. All rights reserved.

T

The spread of the measured values of X4 will be between the upper andlower green curves (except for measurement error). When I> I2, themeasurements will again cluster tightly about the stable equilibria onthe blue curve. The experimenter may not know of the existence of thelimit cycles, but will notice that there is more spread of the measuredvalues of X4 when I1< I< I2. Since the upper and lower green curvesare close to each other, the total set of measured values of X4, over awide range of choices of I, will show the same general shape of a chaircurve with a homeostatic region in the middle. This is what we mean by“homeostasis despite instability.”

In Section 2, we investigate how the phenomenon of homeostasis

despite instability depends on network length, n, properties of thefeedback function, f, and the value of α. In Section 3, we assume that fhas a special form depending on only three parameters, the slope of f,the minimum of f, and the location of the region with constant negativeslope. We present numerical calculations that show how the size of theunstable region and the amplitude of the limit cycles depend on thesethree parameters. In the Discussion, we explain that another homeo-static motif, the parallel inhibition motif [5], shows similar behavior, sohomeostasis despite instability is not limited to the feedback inhibitionmotif.

2. Dependence of stability on network length

Throughout this section we make the following assumptions about f:

> ′ ≤f f fis differentiable, 0, and 0. (1)

We assume f>0 so that the backwards reaction of X2→ X1 does notoccur and the forward reaction of X1→ X2 always occurs at some rate.The assumption that f ′≤ 0 is made so that Xn inhibits the production ofX2 from X1. Additionally, we assume aj>0 for each j. Other hypotheseswill be introduced as appropriate. Let xi(t) denote the concentration ofXi at time t. Each differential equation expresses that the rate of changeof the variable is the rate at which it is made minus the rate at which itis consumed. We assume mass action kinetics for all reactions except forthe rate from X1 to X2 that depends on inhibition from Xn expressedthrough f. The dynamics are given by

= − += −= −⋮= −− −

x I a f x α xx a f x x a xx a x a x

x a x a x

˙ ( ( ) )˙ ( )˙

˙

n

n

n n n n n

1 1 1

2 1 1 2 2

3 2 2 3 3

1 1

where dot indicates a derivative in t. When =α a ,2

+ = − +x x I a x x˙ ˙ ( )1 2 2 1 2 so that as t→∞, + →x x I a/1 2 2. This allows usto reduce the dimension of the steady state Jacobian by 1 and to sig-nificantly simplify the analysis. We are interested in how n and theproperties of f affect the stability of equilibrium solutions to this system.For =n 2, 3, and4, we are able to characterize when an equilibrium isstable, and we can do the same for =n 5 if =α 1. For general n, we givea necessary condition for instability and a sufficient condition for in-stability, but neither is necessary and sufficient.

Using the equations above, it is easy to see that the steady statesolutions satisfy:

Fig. 1. A classic chair curve. The variable Z shows homeostasis and escape fromhomeostasis with respect to the variable I.

Fig. 2. A simple biochemical chain with feedback inhibition. The variable Xn

inhibits the reaction that takes X1 to X2 via the function f(Xn). I is the input tothe chain.

Fig. 3. Homeostasis despite instability. In the feedback chain in Fig. 2, the equilibrium becomes unstable for I in the interval (I1, I2) for an appropriate choice of thefeedback function f. We chose =n 4, =α 1, =a 1j for each j, and = − +− − +f x e x( ) 10 Θ(5 ) 1/2,x1/(5 ) 1/5 where Θ(x) is the Heaviside step function. In Panel A, the blueand red curves show the values of X4(I) at the stable and unstable equilibria, respectively. The green curves show the maximum and minimum values of X4 as thedynamics traverses the limit cycle for fixed I. Since the green curves are close to each other, experimental measurements (see text) will follow the shape of the curveof equilibria despite the instability. In Panel B we show the time course of the oscillations in X4 for =I 10, where the maximum amplitude oscillations are obtained.(For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)

W. Duncan et al. Mathematical Biosciences 300 (2018) 130–137

131

=+

=+

≥

x Ia f x α

xa

a f x Ia f x α

j

( )1 ( )

( )for 2.

n

jj

n

n

11

1

1 (2)

Lemma 1. Suppose f satisfies (1). Then for each I∈ [0, ∞), (2) has aunique solution.

Proof. Let = +h x a f x α a x a f x( ) ( ( ) )( )/( ( ))n1 1 so that =h x I( )n . By thehypotheses on f, h(x) is strictly monotone increasing, =h (0) 0 and h(x)→∞ as x→∞. Thus, h: [0, ∞)→ [0, ∞) is invertible, so that xn isdetermined by I. For j< n, each is an explicit function of xn and I and isthus determined. □

From now on, we drop the overbar and denote =x x I( )j j as thesteady state concentration of Xj. Except in Theorem 1, we consider only

=a 1j for each j. We are interested in the value of xn and the stability ofthe equilibrium. As (2) shows, xn depends only on a1 and an. The valueof an scales xn, so varying an leaves the range of homeostasis unchanged.Changing a1 is equivalent to rescaling f(x), so nothing is lost by setting

=a 11 . We choose =a 1j for = … −j n{2, , 1} for simplicity. Choosingother values makes the proofs more complicated and the statements ofsome inequalities a little different, but involves no new ideas. We willsee later that the choice of α does make a difference.

Differentiating the steady state equation for xn with respect to I andusing (1), we have, for =a 1,j

′ =+

+ − ′x I

f x f x αf x α αf x I

( )( )( ( ) )

( ( ) ) ( )nn n

n n2 (3)

so that ′ >x I( ) 0n for every I. Although ′x I( )n is always nonzero, it maybe small over a range of I so that xn exhibits a homeostatic region. Wewill show in the theorems below that if |f ′(xn)I| is large, then theequilibrium is unstable and if |f ′(xn)I| is small, then the equilibrium isstable. Together with (3), this suggests that the homeostatic region andregion of instability overlap, as depicted in Fig. 3.

The characteristic polynomial of the Jacobian when all =a 1j isgiven by

⎜ ⎟= + + + − ′ ⎛⎝

++

⎞⎠

−P λ f x α λ λ f x I λ αf x α

( ) ( ( ) )(1 ) ( )( )

.n nn

nn

1

(4)

Expanding, this may also be written as

∑= + ⎡⎣⎛⎝

− ⎞⎠

+ + ⎛⎝

−−

⎞⎠⎤⎦

+ ⎡⎣⎢

− + + −′

+⎤⎦⎥

+ ⎡⎣⎢

+ − ′+

⎤⎦⎥

=

−

P λ λ nk

f x α nk

λ

n f x αf x I

f x αλ

f x α f x I αf x α

( ) 1 ( ( ) ) 11

( 1)( ( ) ) 1( )

( )

( ) ( )( )

.

nn

k

n

nk

nn

n

n nn

2

1

Note that every coefficient of Pn is positive (since f ′≤ 0), so that, byDescartes’ rule of signs, there can be no real positive roots of Pn. Inaddition, =λ 0 is never a root, so a loss of stability must occur by a pairof eigenvalues crossing the imaginary axis, that is, via a Hopf bifurca-tion.

When =α 1, = −λ 1 is always a root of Pn(λ). This is a reflection ofthe fact that + =x x I,1 2 mentioned above. Dividing out the factor of

+ λ(1 ) gives us the characteristic polynomial for the reduced Jacobian:

= + + + −′

+∼ −P λ f x λ λ

f x If x

( ) ( ( ) 1 )(1 )( )

( ) 1n nn n

n

2

Expanded, this is

∑= + ⎡⎣⎛⎝

− ⎞⎠

+ + ⎛⎝

−−

⎞⎠⎤⎦

+ +

−′

+

∼ −

=

−

P λ λ nk

f x nk

λ f x

f x If x

( ) 2 ( ( ) 1) 21

( ) 1

( )( ) 1

.

nn

k

n

nk

n

n

n

1

1

2

We first study the case when n≤ 5. Theorems 1 and 2 together showthat =n 4 is the minimum network length for which instabilities canoccur.

Theorem 1. Let f satisfy (1). If =n 2 or 3 and α>0 then the equilibrium islocally asymptotically stable for every I∈ [0, ∞).

Proof. Consider =n 3. Although the theorem is true for general α>0,we prove it in the case =α a2 only. The technique for the general case isthe same as the technique for the = =α n1, 4 case used in the proof ofTheorem 2.

When =n 3 and =α a ,2 using = −x I a x/ ,1 2 2 the reduced Jacobian isgiven by

⎜ ⎟= ⎛⎝

− + ′ −−

⎞⎠

J a f x a a f x I a xa a

( ( ) ) ( )( / )1 3 2 1 3 2 2

2 3

Since − >I a x/ 02 2 and (1) holds, we see <JTr( ) 0 and >Jdet( ) 0 foreach choice of I≥ 0, so that the steady state is asymptotically stable forevery I≥ 0. For the case =n 2, and α>0, the proof is similar. □

Theorem 2. Fix I>0. If =n 4 and α>0, or if =n 5 and =α 1, then fcan be chosen so that f satisfies (1) and the equilibrium is unstable.

Proof. Taking =α 1 simplifies the calculations because it allows us toconsider the roots of ∼P ,n a degree −n( 1) polynomial, when determiningstability rather than the roots of Pn, a degree n polynomial. So, in thecase of =n 4 and α>0 as well as =n 5 and =α 1, we need to localizethe roots of a degree 4 polynomial. We may use the same technique inboth cases. For this reason, the general case of =n 4, α>0 is similar tothe case of = =n α5, 1 and, as in the proof of Theorem 1, we proveTheorem 2 only in the case =α 1.

The main tool of the proof is the argument principle. Consider acontour in the complex plane which connects the points Ri and− Ri viaa semi-circle in the right half plane and a line on the imaginary axis. Forlarge R, the dominant term of ∼Pn on the semi-circle is of the form

− −R en n iθ( 1) ( 1) . So, in the limit as R→∞, the change in the argument onthe semi-circle is −n π( 1) . For �∈z , let ℜ(z) denote the real part of zand ℑ(z) denote the imaginary part. We may determine the change inthe argument on the imaginary axis between Ri and − Ri by computingthe zeros and tracking the signs of ∼P iy( ( ))nR and ∼P iy( ( ))nI . For =n 4, thezeros of ∼P iy( ( ))4R satisfy

=+ − ′+ +

= >yf x f x If x f x

r( ( ) 1) ( )( ( ) 3)( ( ) 1)

: 0Re2 4

24

4 4

and the zeros of ∼P iy( ( ))4I satisfy

= = + = >y y f x r0 or 2 ( ) 3 : 0.Im2

4

To compute the change in the argument of on the imaginary axis, wetrack which quadrant of the complex plane ∼P iy( )4 lies in as y varies from∞ to − ∞. Note that = − − +∼P iy C y r y r( ( )) ( )( )Re Re4R where C is apositive constant and = − − +∼P iy y y r y r( ( )) ( )( )Im Im4I . For y large,

<∼P iy( ( )) 04R and <∼P iy( ( )) 04I so ∼P iy( )4 lies in quadrant 3. ∼P iy( ( ))4R and∼P iy( ( ))4I have only simple roots, so the quadrant changes exactly when

y passes through one of the roots. The path ∼P iy( )4 takes, and thereforethe change in the argument, depends on whether rRe< rIm or rIm< rRe.We consider these two cases separately.

Case 1: rIm< rReThe number line below shows which quadrant of the complex plane

∼P iy( )4 lies in.∼P iy( )4 goes from − ∞i to i∞ as y goes from ∞ to − ∞ and, as the

number line shows, makes no additional revolutions in between. Thatis, the change in the argument of on the imaginary axis is π. Along with

W. Duncan et al. Mathematical Biosciences 300 (2018) 130–137

132

the change of 3π on the arc of the semi-circle, the total change in theargument of on the whole contour is 4π, indicating that there are 2roots with positive real part when rIm< rRe.

Case 2: rRe< rIm

The change in the argument of on the imaginary axis is − π3 in thiscase. The total change on the whole contour is then 0 and there are noroots inside the contour.

Let λ *4 be a root of P4 with largest real part. The analysis aboveshows that = −λ r rsign( ( *)) sign( )Re Im4R . Simplifying this expression,we have

= − ′ − + +λ f x I f x f xsign( ( *)) sign( ( ) (2( ( ) 1)( ( ) 2) )4 4 4 42R (5)

For =n 5 and =α 1 we again compute the zeros of the real andimaginary parts of ∼Pn in order to determine which quadrant it lies in as ittraverses the imaginary axis. The zeros of ∼P iy( ( ))5R satisfy

⎜ ⎟= + ± ⎡⎣⎢ + − ⎛

⎝+ −

′+

⎞⎠⎤⎦⎥

= ±

y f x f x f xf x I

f x

r

2 (3 ( ) 6) (3 ( ) 6) 4 ( ) 1( )

( ) 1

:2

25 5

25

5

5

1/2

and the zeros of ∼P iy( ( ))5I satisfy

= =++

=y yf xf x

r0 or3 ( ) 4

( ) 4: Im

2 5

5

It can be shown that the change in the argument of ∼P iy( )5 as y goes from∞ to − ∞ is − π4 if and only if �∈±r and < <− +r r rIm . Further, if

�∈±r , then < +r rIm is always satisfied. Let λ *5 be a root of P5 withlargest real part. Then, sign(λ *5 ) ≥ sign( −−r rIm) which is equal to thesign of

⎜ ⎟

− ′ − + ⎡

⎣⎢

+ ++

− +

− ⎛⎝

++

⎞⎠⎤

⎦⎥

f x I f xf x f x

f xf x

f xf x

4 ( ) ( ( ) 1)4(3 ( ) 4)(3 ( ) 6)

( ) 44( ( ) 1)

6 ( ) 8( ) 4

.

5 55

55

5

5

2

(6)

Now, fix I>0 and pick any f satisfying (1). The equilibrium value isindependent of network length, so let = =x x x4 5. Now, if we have

>λsign( ( *)) 04R and >λsign( ( *)) 05R then the equilibrium is unstableand we are done. If not, then we may choose g satisfying (1) with

=g x f x( ) ( ) so that the equilibrium value with g as the feedbackfunction is the same, but with ′g x( ) large enough so that

>λsign( ( *)) 04R and >λsign( ( *)) 0,5R which guarantees the equili-brium is unstable. □

Theorem 3. Let f satisfy (1), I>0, and =α 1. Let λ *4 and λ *5 be the rootsof and ∼P5 with largest real part. If ≥λ( *) 04R then >λ( *) 05R .

Note that in particular, Theorem 3 says that for =α 1, if the networkwith length 4 is unstable, then the network with length 5 is unstable aswell.

Proof of Theorem 3. The steady state value of xn is independent of n.So, the theorem is proved if the expression in (6) is strictly larger thanthe expression in (5). Letting = =x x x ,4 5 this requires

⎜ ⎟

− ′ − + ⎡

⎣⎢

+ ++

− +

− ⎛⎝

++

⎞⎠⎤

⎦⎥

> − ′ − + +

f x I f xf x f x

f xf x

f xf x

f x I f x f x

4 ( ) ( ( ) 1)4(3 ( ) 4)(3 ( ) 6)

( ) 44( ( ) 1)

6 ( ) 8( ) 4

( ) 2( ( ) 1)( ( ) 2)

2

2

Rearranging the inequality, we see this is true if and only if

⎜ ⎟− ⎛⎝

++

⎞⎠

<+

+ + +f xf x f x

f x f x f x6 ( ) 8

( ) 44

( ) 4[2 ( ) 8 ( ) 15 ( ) 12]

23 2

The left hand side is negative and the right hand side is positive sothe inequality is always satisfied. □

We now study stability in the case of general n. By applyingGershgorin’s circle theorem (see, for example, [7]) to the Jacobian, wefind a necessary condition for instability of the equilibrium, which ispresented in the following proposition.

Proposition 1. Let f satisfy (1) and fix I. Suppose the equilibrium isunstable. If =α 1 then − ′ > +f x I f x( ) ( ) 1n n . If α≠ 1 then− ′ > +f x I f x α2 ( ) ( )n n .

The following theorem provides a sufficient condition for in-stability.

Theorem 4. Let f satisfy (1) and fix I, n≥ 4, and α>0. Suppose that

− ′ > + −

−( )( )f x I f x α( ) 2( ( ) ) secn nπ

n

n32( 1)

1and f(xn)< α. Then for

m≥ n, the equilibrium of the network with length m is unstable.

Proof. We use notation and Theorems 1 and 2 of [8]. First, consider thenetwork of length n. Let = −Q λ P λ( ) ( 1)n n where Pn defined as in (4). Pnhas a root with positive real part when Qn has a root with real partgreater than 1. Define =− ′

+γ: f x If x α

( )( )

nn

. We may write

= − + + +− −Q λ λ α λ γ f x λ( ) ( 1 )( ) ( )nn

nn1 1

and view Qn(λ) as a perturbation of = − + +−p λ λ α λ γ( ): ( 1 )( )n 1 . Aroot of p with largest real part is given by

= − −λ γ e*: .n iπ n1/( 1) /( 1)

Let Z(p, ε) be the root neighborhoods of p under the metric= −d p q a b m( , ) max /j j j with =−m 1n 1 and =m 0j for ≠ −j n 1 as

defined in [8]. Let Z*(p, ε) be the connected component of Z(p, ε)containing λ*. By Theorem 2 of [8] there is at least one root of Qn inZ*(p, f(xn)). So it is sufficient to show that

W. Duncan et al. Mathematical Biosciences 300 (2018) 130–137

133

�

⎜ ⎟

⊂ =⎧⎨⎩

∈ >

∈ ⎛⎝ − −

⎞⎠⎫⎬⎭

Z p f x z z z

πn

πn

* ( , ( )) Ω: : ( ) 1 and arg( )

2( 1), 3

2( 1).

n R

First, we show that λ*∈Ω.

⎜ ⎟⎜ ⎟

= ⎛⎝ −

⎞⎠

> ⎛⎝

⎛⎝ −

⎞⎠⎞⎠

⎛⎝ −

⎞⎠

> ⎛⎝ −

⎞⎠≥ ≥

−

−

λ γ πn

πn

πn

πn

n

( *) cos1

2 sec 32( 1)

cos1

2 cos1

1 for 4

n

n

1/( 1)

1

R

We also have = −λarg( *) πn2( 1) and thus λ*∈Ω. As Z*(p, f(xn)) is

connected, the above containment holds if ∂ ∩ = ⌀Z p f xΩ ( , ( ))n . Let= −g z p z z( ) ( )/ n 1. By Theorem 1 of [8], z∈ Z(p, f(xn)) if and only if |g

(z)|≤ f(xn).

Now suppose z∈ ∂Ω with =z( ) 1R and ∈ − −( )zarg( ) ,πn

πn2( 1)3

2( 1) .We have

=− + +−

−g zz α z γ

z( )

1.

n

n

1

1

=z( ) 1R so − =z z i1 ( )I and thus − + >z α α1 . Using the reversetriangle inequality we have

⎜ ⎟⎜ ⎟

> −

≥ − ⎛⎝

⎛⎝ −

⎞⎠⎞⎠

> − = >

− −

− −

g z α z γ

α πn

γ

α α f x

( ) 1

1 sec 32( 1)

1 2 ( )

n

n

n

( 1)

( 1)

and we have z∉ Z(p, f(xn)).

Now suppose = −zarg( ) πn2( 1) and ℜ(z)≥ 1. Then = −z re

πin2( 1) for

⎟∈ ⎡⎣⎢

∞⎞⎠

−( )r sec ,πn2( 1) and

⎜ ⎟

⎜ ⎟

⎛⎝

⎞⎠

= − + +

> − +

> ⎛⎝ −

⎞⎠− + ≥ >

− − −− −

−

g re r re α r i γ

re α

r πn

α α f x

1

1

cos2( 1)

1 ( )

n n

n

( 1) 1πi

nπi

n

πin

2( 1) 2( 1)

2( 1)

so z∉ Z(p, f(xn)).

Finally, suppose = −zarg( ) πn3

2( 1) and ℜ(z)≥ 1. Then = −z reπi

n3

2( 1) for

⎟∈ ⎡⎣⎢

∞⎞⎠

−( )r sec ,πn3

2( 1) and

⎜ ⎟

⎜ ⎟

⎛⎝

⎞⎠

= − + − +

> − +

> ⎛⎝ −

⎞⎠− + ≥ >

− − −− −

−

g re r re α ir γ

re α

r πn

α α f x

1

1

cos 32( 1)

1 ( )

n n

n

( 1) 1πi

nπi

n

πin

32( 1)

32( 1)

32( 1)

so again, z∉ Z(p, f(xn)) and we have shown the case of =m n. For

m> n, note that −

−( )( )sec πn

n32( 1)

1is decreasing in n so that if the hy-

potheses are satisfied for =m n, they must also be satisfied form> n. □

In practice, if we are given f, n, and α and we wish to know if there isan I for which the equilibrium is unstable, we may use (2) to solve for Ias a function of xn: = +I x f x α f x( ( ) )/ ( )n n n . Then the hypotheses ofProposition 1 and Theorem 4 can be written independent of I.Proposition 1 can then be used to find xn for which stability is guar-anteed and Theorem 4 can be used to find xn for which instability isguaranteed. The corresponding values of I can be found by substitutingxn back into (2) (Fig. 4).

Fig. 4. Instability regions as a function of network length. A: The solid red area shows the instability region with = ++

f x( ) 1/4x

101 10 and =α 1

50 computed nu-

merically. f(x)> α for all x so Theorem 4 does not hold and if n is large enough the equilibrium is stable for every I∈ [0, ∞). Outside of the green lines, Proposition 1applies so the equilibrium must be stable. B: f(x) as in A but =α 1 so that there are I for which Theorem 4 applies. The dark striped area shows where Theorem 4guarantees instability and the solid red area shows where we have numerically calculated instabilities to occur. (For interpretation of the references to color in thisfigure legend, the reader is referred to the web version of this article.)

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3. A special case

To better understand how the properties of f affect the system, weconsider a particular form of f when =n 4. Consider f of the form

=⎧⎨⎩

+ <+ − ≤ ≤

>f x

A C xA C sx sx AC sx A

( ), 0

, 0, (7)

where A> C, s>0, and C≥ 0. Let = −f x f x h( ) ( )h for h≥ 0. We willexamine how varying C, s, and h effect the instability region and am-plitude of limit cycles when =n 4 and =α 1. We provide two theorems.To state the theorems, we define = >I I x Iinf{ 0: ( )isunstable}1 and= >I I x Isup{ 0: ( )isunstable}2 and note that (I1, I2) is the instability

region. The first theorem relates the parameters of fh with the existenceof the instability region. The second theorem shows that when the in-stability region is not empty, −x I x I( ) ( )4 2 4 1 must be relatively small.The purpose of the second theorem is to show that the region of in-stability is contained in the homeostatic region.

Theorem 5. Define fh(x) as above, let =n 4 and =α 1. Let C*>0 satisfy+ = +C C A sh2 *( * 2)2 . If C> C* then the equilibrium is stable for every

I∈ [0, ∞). If 0< C< C*, then the instability region is nonempty andI2<∞. If =C 0 then the instability region is nonempty and = ∞I2 .

Proof. First note that if x4< h or − >s x h A( )4 then ′ =f x( ) 0h 4 and byProposition 1, the equilibrium is stable.

= − + − − + + − −g x sx A C s x h A C s x hLet ( ) 2( ( ))( 2 ( )) .2

For ∈ +x h A s h( , / ),4 applying Eq. (5) from the proof of Theorem 2shows that the equilibrium is stable if g(x4)< 0 and unstable if g(x4)> 0. Note that g is strictly monotone increasing and

+ = + − +g A s h A sh C C( / ) ( ) 2 (2 )2.Suppose C> C*. Then < + <g x g A s h( ) ( / ) 0 for all

∈ +x h A s h( , / ) so that the equilibrium is stable for every I≥ 0.Now suppose C< C*. Then + >g A s h( / ) 0 so there is an interval

⊂ +U h A s h( , / ) so that if x4∈U then g(x4)> 0 and the equilibrium isunstable. If, in addition, C>0, then the proof of Lemma 1 still worksand x4→∞ as I→∞ so that there is an I for which > +x A s h/4 and

stability is regained. In this case I2 satisfies = +x I A s h( ) /4 2 .On the other hand, if =C 0, then as in Lemma 1, x4 satisfies

= + =x f x x f x Iℓ( ): ( ( ) 1) / ( )4 4 4 4 . Because f(x)→ 0 as → +x A s h/ , wesee that ℓ(x)→∞ as → +x A s h/ and + → ∞A s hℓ: [0, / ) [0, ) is in-vertible. Thus we always have < +x A s h/ ,4 so stability is not regainedand = ∞I2 . □

Theorem 6. Define fh(x) as above, let =n 4 and =α 1. Suppose theinstability region is nonempty. Then >x I x I( ) ( )4 1

89 4 2 .

Proof.

= − + − − + + − −g x C sx A C s x h A C s x hDefine ( , ) 2( ( ))( 2 ( )) .2

This is the same g as in the proof of Theorem 5 except that we make thedependence on C explicit. We know that the equilibrium is stable wheng<0 and unstable when g>0. Note that g is decreasing in C, so wehave g(x, C)< g(x, 0). We also have = +x I A s h( ) /4 2 . Letting =κ ,1

9 wehave

⎛⎝

⎞⎠

= − +

= − + − − + + +≤ − − += − − − −< − =

g x I g κ A s h

κ A κ sh κA κsh κA κshκ A κA κAκ A κA κA κA

A κA

89

( ), 0 ((1 )( / ), 0)

(1 ) (1 ) 2( )( 2)(1 ) 2 ( 2)(1 ) 2( ) 8 8

9 0

4 2

2

2

3 2

The instability region is nonempty so I1 exists and x4(I1) satisfies=g x I C( ( ), ) 04 1 . However, g is increasing in x and we have shown that<g x I C( ( ), ) 08

9 4 2 so we must have ∈ ( )x I x I x I( ) ( ), ( )4 189 4 2 4 2 as

claimed. □

Theorem 6 says that although −I I2 1 may be large, the value of x4doesn’t change very much on that interval. This indicates that the in-stability region lies on the homeostatic plateau. For example, Fig. 5Ashows that for =C 1/5, I1≈ 25 while I2≈ 100, but x4(I1) and x4(I2) arestill close to each other.

Finally, we present numerical calculations of the instability regionsand limit cycle amplitudes as C, s, and h are individually varied. Fig. 5shows the instability regions. Fig. 6 shows the maximum limit cycle

Fig. 5. Parameter dependence of the instability region. fh(x) is chosen as above with =A 10, =C 1/5, =s 1, and =h 0 except when they are varied. The solid redregions show the numerically calculated instability regions. A: When =A 10 and =h 0, C*≈ 0.69. As C*→ 0, I2→∞. B: Increasing s decreases both I1 and I2. Theline =s 0.96 has been plotted and indicates for which values of I the equilibrium is stable (blue), and for which values of I there is a limit cycle (green). I1 and I2 arethe intersection points of the line =s 0.96 and the boundary of the instability region as indicated. For small s, +A s h/ is very large, causing I2 to be very large. C:Increasing h increases the size of the instability region. (For interpretation of the references to color in this figure legend, the reader is referred to the web version ofthis article.)

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amplitude over the instability region. We choose A large enough so thatx4(I1)> h.

4. Discussion

Gene expression levels vary by roughly 25% from individual to in-dividual [9,10]. Furthermore, many genes that code for enzymes havepolymorphisms with high frequency in the human population that arefunctional in the sense that they raise or lower the activity of the en-zyme by 20% to 80% (see, for example, Table 1 in [11]). How do cel-lular and physiological systems cope with this enormous biologicalvariation? The answer is that there is a myriad of homeostatic me-chanisms that reduce the effects of gene expression differences and genepolymorphisms. Some of these mechanisms operate at the gene level,some operate at the biochemical level, and some at the physiologicallevel. But many also operate between levels, because biochemicalsubstrates affect gene expression levels as do physiological variablessuch as hormones. These homeostatic regulations should not be thought

of as a separate layer of regulation since they probably evolved si-multaneously with the genetic, biochemical, and physiological systems.They are integral parts of the whole system and must be investigated inorder to understand cell biology and whole body physiology.

It is well known that feedback inhibition with explicit time delayscan produce oscillations in mechanical or biological systems. In the caseof feedback inhibition in a biochemical chain, there is no explicit timedelay, but a time delay is implicit in the length of the chain, n. As wehave shown, the existence of an input interval (I1, I2) in which there areoscillations, the length of the interval, and the size of the oscillationsdepends in interesting and complicated ways on the properties of theinhibition function, f, the length of the chain, and the size of the leakageparameter, α. Our main point is that when the amplitudes of the os-cillations are relatively small, one maintains the homeostatic plateauand the chair shape despite the instability of the equilibrium and thelimit cycle oscillations.

This phenomenon is not confined to the homeostatic mechanism offeedback inhibition. In [5] (Fig. 9), we introduced the parallel

Fig. 6. Parameter dependence of the limit cycle amplitudes. The maximum amplitude of the limit cycles over the instability region is calculated numerically andplotted as a function of the parameters of fh. The parameters are fixed at =A 10, =s 1, =C 1/5, and =h 0 when they are not varied. The maximum amplitudesappear to be proportional to the size of the instability region. By amplitude, we mean the minimum of x4 subtracted from the maximum over the limit cycle.

Fig. 7. Homeostasis despite instability in the parallel inhibition motif. The network is shown in Panel A. All reactions are linear mass action except for the twoinhibitions given by f and g. Panel B shows the instability region in red and the maximum amplitude of the limit cycle oscillations in green. The amplitude of theoscillations can be adjusted by changing parameters. The differential equations and the explicit formulas for f and g are available from the authors. (For interpretationof the references to color in this figure legend, the reader is referred to the web version of this article.)

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inhibition motif and showed that it can produce the classic chair shapewith a homeostatic plateau and escape from homeostasis. It is easy toprove that the two-variable implementation of the parallel inhibitionmotif given in [5] has a stable equilibrium for all inputs I. However,consider the four-variable implementation of the parallel inhibitionmotif given in Fig. 7, Panel A. Each of the inhibitions acts through anintermediate variable (Y or W), and, as can be seen in Panel B, thehomeostatic variable, Z, shows the phenomenon we study in this paper.For more discussion of the parallel inhibition motif and an explicitbiological example, see [5].

This paper introduces a mathematically new low codimensionphenomenon in network dynamics - the coexistence of Hopf bifurcationwith homeostasis. Both of these singularities occur on variation of I,which simultaneously serves the roles of input parameter and bifurca-tion parameter. Infinitesimal chairs [6] have codimension one and theco-existence with Hopf bifurcation has codimension two. However, theexistence of codimension two phenomena should not be surprising inbiochemical networks which have many parameters (rate constants andthe like). Note that Fig. 3 suggests a degenerate Hopf bifurcation[12,13] where a pair of Hopf bifurcations merge and disappear coupledwith an infinitesimal chair, which is a codimension three phenomenon.Theoretically, the coexistence of homeostasis with bifurcation is a fas-cinating mathematical problem that should lead to new and interestingnetwork issues. For example, another codimension three interaction isthat of a hysteresis point [13] with a chair singularity. Such an inter-action might lead to the coexistence of two plateaus with the possibilityof a new form of switching. In any case, understanding the theoreticalstructure of these mathematical interactions will be a subject of futurestudy.

Competing interest

The authors declare that they have no competing interests.

Authors’ contributions

The phenomenon of homeostasis despite instability was discovered

by Best and the mathematical calculations were done by Duncan.Nijhout and Reed organized the project and Golubitsky contributed theperspective of singularity theory. The paper was written by Best,Duncan, and Reed, and was read and approved by all authors.

Acknowledgements

This research was supported by National Institutes of Health grants1R01MH106563-01A1(JAB,MCR,HFN) and 1R21MH109959-01A1(JAB,MCR,HFN) and NSF grants IOS-1562701 (HFN), EF-1038593(HFN,MCR), IOS-1557341 (HFN,MCR).

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